 Cubic nilpotency from a mixed chiral constraint

On the invertible-$F_X$ branch, use $y=(G\chi )/F_X-(GG)F_Y/(2F_X^2)$ and the two-component Grassmann algebra identity $(G\chi {)}^2=-\frac{1}{2}(GG)(\chi \chi )$. Then $y^2=-(GG)(\chi \chi )/(2F_X^2)$, $y^3=y^2\chi =0$ and $y^2{F}_Y-y\chi \chi =0$. These are the components of $Y^3=0$. Generally $y^2$ is nonzero, so cubic nilpotency must not be replaced by quadratic nilpotency. The allowed analytic holomorphic monomials are $1,X,Y,Y^2$.